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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Mechanical Properties of Solids

JIPMER 2018 · Physics · Mechanical Properties of Solids

If compressibility of material is 4 × 10^ -5 per atm, pressure is...

If compressibility of material is 4×10−54 \times 10^{-5} per atm, pressure is 100 atm and volume is 100 cm3100 \mathrm{~cm}^3, then find ΔV=\Delta \mathrm{V}= ?

Options

  1. A0.4 cm30.4 \mathrm{~cm}^3
  2. B0.8 cm30.8 \mathrm{~cm}^3
  3. C0.6 cm30.6 \mathrm{~cm}^3
  4. D0.2 cm30.2 \mathrm{~cm}^3

Correct answer

A. 0.4 cm30.4 \mathrm{~cm}^3

Step-by-step solution

∵\because Compressibility ∣β∣=1∣ Bulk modulus of  elasticits ∣∣β∣=ΔpVΔVorβ=l(ΔpVΔV)=ΔVΔpV|\beta|=\frac{1}{\mid \text{ Bulk modulus\text{ of } elasticits } \mid} |\beta|=\frac{\Delta \mathrm{pV}}{\Delta \mathrm{V}} or \quad \beta=\frac{l}{\left(\frac{\Delta \mathrm{pV}}{\Delta \mathrm{V}}\right)}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{pV}}∴ΔV=β(Δp.V)=4×10−5(100×100×10−6)=0.4×10−6m3=0.4cm3\begin{aligned} \therefore \Delta \mathrm{V}=\beta(\Delta \mathrm{p} . \mathrm{V}) & =4 \times 10^{-5}\left(100 \times 100 \times 10^{-6}\right) \\ & =0.4 \times 10^{-6} \mathrm{ m}^3=0.4 \mathrm{ cm}^3\end{aligned}$$
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